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When a ball is dropped from a state of rest at time t = 0, the distance it has traveled after t seconds is s(t) = 16t2. (a) How far does the ball travel during the time interval [4, 4.5]? 68 ft (b) Compute the average velocity over the time interval [4, 4.5]. ft/sec (c) By computing the average velocity over the time intervals [4, 4.100], [4, 4.010], and [4, 4.001], . . . , estimate the ball's instantaneous velocity at t = 4. (Round your answer to one decimal place.) ft/sec.

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Answer:

Part a)

[tex]d = 68 ft[/tex]

Part b)

[tex]v_{avg} = 136 ft/s[/tex]

Part c)

[tex]v = 128 ft/s[/tex]

Explanation:

Part a)

As we know that displacement is given as

[tex]s = 16 t^2[/tex]

now the displacement in 4.5 s and t = 4 s is given as

[tex]y_1 = 16(4.5)^2 = 324 ft[/tex]

[tex]y_2 = 16(4^2) = 256 ft[/tex]

now displacement is given as

[tex]d = y_1 - y_2[/tex]

[tex]d = 324 - 256[/tex]

[tex]d = 68 ft[/tex]

Part b)

So displacement of the ball in given interval of t = 4.00 to 4.5 s

[tex]d = 68 ft[/tex]

time interval is given as

[tex]\Delta t = 0.5 s[/tex]

average velocity is given as

[tex]v_{avg} = \frac{68}{0.5}[/tex]

[tex]v_{avg} = 136 ft/s[/tex]

Part c)

So displacement of the ball in given interval of t = 4.00 to 4.1 s

[tex]d = 16(4.1^2 - 4^2)[/tex]

[tex]d = 12.96 ft[/tex]

average velocity is given as

[tex]v_{avg} = \frac{12.96}{0.1}[/tex]

[tex]v_{avg} = 129.6 ft/s[/tex]

Similarly for time interval t = 4.0 s to t = 4.01 s

[tex]d = 16(4.01^2 - 4^2)[/tex]

[tex]d = 1.2816 ft[/tex]

average velocity is given as

[tex]v_{avg} = \frac{1.2816}{0.01}[/tex]

[tex]v_{avg} = 128.16 ft/s[/tex]

Similarly for time interval t = 4.0 s to t = 4.001 s

[tex]d = 16(4.001^2 - 4^2)[/tex]

[tex]d = 0.128016 ft[/tex]

average velocity is given as

[tex]v_{avg} = \frac{0.128016}{0.001}[/tex]

[tex]v_{avg} = 128.016 ft/s[/tex]

so instantaneous velocity is given as

[tex]v = 128 ft/s[/tex]

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